我推荐这个解决方案。您需要继承和重写方法运行。

import sys
import os
from signal import SIGTERM
from abc import ABCMeta, abstractmethod



class Daemon(object):
    __metaclass__ = ABCMeta


    def __init__(self, pidfile):
        self._pidfile = pidfile


    @abstractmethod
    def run(self):
        pass


    def _daemonize(self):
        # decouple threads
        pid = os.fork()

        # stop first thread
        if pid > 0:
            sys.exit(0)

        # write pid into a pidfile
        with open(self._pidfile, 'w') as f:
            print >> f, os.getpid()


    def start(self):
        # if daemon is started throw an error
        if os.path.exists(self._pidfile):
            raise Exception("Daemon is already started")

        # create and switch to daemon thread
        self._daemonize()

        # run the body of the daemon
        self.run()


    def stop(self):
        # check the pidfile existing
        if os.path.exists(self._pidfile):
            # read pid from the file
            with open(self._pidfile, 'r') as f:
                pid = int(f.read().strip())

            # remove the pidfile
            os.remove(self._pidfile)

            # kill daemon
            os.kill(pid, SIGTERM)

        else:
            raise Exception("Daemon is not started")


    def restart(self):
        self.stop()
        self.start()

我推荐这个解决方案。您需要继承和重写方法运行。

import sys
import os
from signal import SIGTERM
from abc import ABCMeta, abstractmethod



class Daemon(object):
    __metaclass__ = ABCMeta


    def __init__(self, pidfile):
        self._pidfile = pidfile


    @abstractmethod
    def run(self):
        pass


    def _daemonize(self):
        # decouple threads
        pid = os.fork()

        # stop first thread
        if pid > 0:
            sys.exit(0)

        # write pid into a pidfile
        with open(self._pidfile, 'w') as f:
            print >> f, os.getpid()


    def start(self):
        # if daemon is started throw an error
        if os.path.exists(self._pidfile):
            raise Exception("Daemon is already started")

        # create and switch to daemon thread
        self._daemonize()

        # run the body of the daemon
        self.run()


    def stop(self):
        # check the pidfile existing
        if os.path.exists(self._pidfile):
            # read pid from the file
            with open(self._pidfile, 'r') as f:
                pid = int(f.read().strip())

            # remove the pidfile
            os.remove(self._pidfile)

            # kill daemon
            os.kill(pid, SIGTERM)

        else:
            raise Exception("Daemon is not started")


    def restart(self):
        self.stop()
        self.start()

使用你的系统提供的任何服务管理器——例如在Ubuntu下使用upstart。这将为您处理所有细节，如启动时启动，崩溃时重新启动等。

cron is clearly a great choice for many purposes. However it doesn't create a service or daemon as you requested in the OP. cron just runs jobs periodically (meaning the job starts and stops), and no more often than once / minute. There are issues with cron -- for example, if a prior instance of your script is still running the next time the cron schedule comes around and launches a new instance, is that OK? cron doesn't handle dependencies; it just tries to start a job when the schedule says to.

如果您发现确实需要一个守护进程(一个永不停止运行的进程)，请查看一下监控器。它提供了一种简单的方法来包装普通的、非守护进程化的脚本或程序，并使其像守护进程一样运行。这比创建本地Python守护进程好得多。

如果你正在使用终端(ssh或其他东西)，并且你想在退出终端后保持长时间的脚本工作，你可以尝试这样做:

屏幕

Apt-get安装屏幕

在里面创建一个虚拟终端(即abc): screen -dmS abc

现在我们连接到abc: screen -r abc

因此，现在我们可以运行python脚本:python keep_sending_mail .py

从现在开始，你可以直接关闭你的终端，但是，python脚本将继续运行而不是被关闭

因为这个keep_sending_mail .py的PID是虚拟屏幕的子进程，而不是 终端(ssh)

如果你想检查你的脚本运行状态，你可以再次使用screen -r abc

您可以在脚本中或像这样在另一个脚本中作为子进程运行进程

subprocess.Popen(arguments, close_fds=True, stdout=subprocess.DEVNULL, stderr=subprocess.DEVNULL, stdin=subprocess.DEVNULL)

或者使用现成的工具

https://github.com/megashchik/d-handler